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V_{R} + V_{L} + V_{C} = 0 V is o.k.
WIth the current direction as shown this directly translates to
[math] i \times R + L \times \frac{di}{dt} + \frac{q}{C} = 0 V[/math]I don't know why you use negative values in your equation, but this is a mere formal complaint as both your and my equation are equivalent since the result is 0 V and in your 4th line on the left you miraculously come tothe samea similar result, but why do you change the integral from [math]\int i dt[/math] to [math]\int v dt[/math]? This is wrong.
Coming to the right side: you make matters unnecessarily complicated by distinguishing I_{R}, I_{L} and I_{C}. Since this circuit has only one loop, I_{R} = I_{L} = I_{C} = I simplifies writing down the equations. You furthermore make it very difficult to follow your writeup when you e.g. [math]V = L \times \frac{di}{dt}[/math] without using the subscript "L" for V_{L}.
Please correct your writeup so we can follow your reasoning without headaches.
By the way: It is imho good practice to keep the units into the calculations. This allows to check the result: If the units don't match there is definitely an error in the calculation.
but the question is "determine the characteristic equation for the voltage across the capacitor (iC), the voltage across the inductor (iL) and the voltage across the resistor (iR)." did I just solved the problem asked?Looks o.k up to this point:
View attachment 61345
The next 2 equations are imho not helpful.
The equation now has the form x''+ax'+bx = 0 with a = R/L and b = 1/(LC). This second order differential equation can now be solved using with some help from the internet, see e.g. this page. Once you have the solution, you can resubstitute a and b by the respective expressions.
Btw: Your thread headers are not very informative. Instead of "help meeee I have my answer but I'm not sure" something like "help me solve an LRcircuit" immediately gives our members a hint what your question is about.